4x4

4! * 4 - 4/.4 = 86

(!4)*(4/.4) - 4 = 86

Originally posted by Agerg
(!4)^(4 - sqrt(4)) + sqrt(4) = 83

!4^sqrt(4) + 4 + sqrt(4) = 87

44 * (4/4) = 88

arccos(4 - 4) - 4/4 = 89

hmm...I think perhaps using !n is cheaper than inverse trig operations because for example I can find 1 via : !(!(sqrt(!4)))...I'm betting with a bit of creativity one could find many integers with just one 4 and a combination of sqrt(n), n!, and !n 😵

4 * 4! - 4 - sqrt(4) = 90

Originally posted by Agerg
hmm...I think perhaps using !n is cheaper than inverse trig operations because for example I can find 1 via : !(!(sqrt(!4)))...I'm betting with a bit of creativity one could find many integers with just one 4 😵

arccos(4 - 4) + 4/4 = 91

!4 * 4 * .4 + sqrt(2) = 92

But the thing is to use exactly four 4's. I think it's good enough. But if others think so, we can continue without it.

true...but finding a three now costs only one 4 where as before it would cost you 3 of em 🙂

Originally posted by kbaumen
But the thing is to use exactly four 4's. I think it's good enough. But if others think so, we can continue without it.

I figured that you can use also 4'. Derivative from any constant number is 0 so figure it out with less than four 4's and add a number of 4' you need.

For example 4/.4 + 4' + 4' = 10

Originally posted by kbaumen
I figured that you can use also 4'. Derivative from any constant number is 0 so figure it out with three 4's and add 4'.